Binary Tree Inorder Traversal LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Binary Tree Inorder Traversal in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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• Mathematics/Basic Logical Based Questions
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• Database
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• Recursion, etc.

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Link for the Problem – Binary Tree Inorder Traversal– LeetCode Problem

Binary Tree Inorder Traversal– LeetCode Problem

Problem:

Given the root of a binary tree, return the inorder traversal of its nodes’ values.

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Binary Tree Inorder Traversal– LeetCode Solutions

Binary Tree Inorder Traversal Solution in C++:

class Solution {
 public:
  vector<int> inorderTraversal(TreeNode* root) {
    vector<int> ans;
    stack<TreeNode*> stack;

    while (root || !stack.empty()) {
      while (root) {
        stack.push(root);
        root = root->left;
      }
      root = stack.top(), stack.pop();
      ans.push_back(root->val);
      root = root->right;
    }

    return ans;
  }
};

Binary Tree Inorder Traversal Solution in Java:

class Solution {
  public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> ans = new ArrayList<>();
    Stack<TreeNode> stack = new Stack<>();

    while (root != null || !stack.isEmpty()) {
      while (root != null) {
        stack.push(root);
        root = root.left;
      }
      root = stack.pop();
      ans.add(root.val);
      root = root.right;
    }

    return ans;
  }
}

Binary Tree Inorder Traversal Solution in Python:

class Solution:
  def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
    ans = []
    stack = []

    while root or stack:
      while root:
        stack.append(root)
        root = root.left
      root = stack.pop()
      ans.append(root.val)
      root = root.right

    return ans